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a. \(\dfrac{-3}{5}\)
b. \(\dfrac{-2}{3}\) c. \(\dfrac{4}{39}\) d. \(\dfrac{26}{45}\)
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)
\(\left(6-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{18}{5}.\dfrac{1}{4}\)
\(=\left(\dfrac{30}{5}-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{18}{5}.\dfrac{1}{4}\)
\(=\dfrac{16}{5}.\dfrac{25}{8}-\dfrac{18}{20}\)
\(=\dfrac{16.25}{5.8}-\dfrac{18}{20}\)
\(=\dfrac{2.5}{1}-\dfrac{9}{10}\)
\(=10-\dfrac{9}{10}\)
\(=\dfrac{100}{10}-\dfrac{9}{10}\)
\(=\dfrac{91}{10}=9\dfrac{1}{10}\)
\(\left(6-\dfrac{14}{5}\right)\) . \(\dfrac{25}{8}\) - \(\dfrac{18}{5}\) . \(\dfrac{1}{4}\)
=\(\left(\dfrac{30}{5}-\dfrac{14}{5}\right)\) . \(\dfrac{25}{8}\) - \(\dfrac{18}{5}\) .\(\dfrac{1}{4}\)
= \(\dfrac{16}{5}\) . \(\dfrac{25}{8}\) - \(\dfrac{9}{10}\)
= \(\dfrac{16.25}{5.8}\) - \(\dfrac{9}{10}\)
= \(\dfrac{2.5}{1}\) - \(\dfrac{9}{10}\)
= 10 -\(\dfrac{9}{10}\)
= \(\dfrac{100}{10}\) - \(\dfrac{9}{10}\)
= \(\dfrac{91}{10}\) = 9,1
mình ghi nhầm nên các bạn cứ hết hai phân số là một câu nhé ví dụ như \(\dfrac{-5}{8}\):\(\dfrac{15}{4}\)
a )
\(\frac{2}{7}+\frac{7}{7}.\frac{14}{25}\)
\(=\frac{2}{7}+1.\frac{14}{25}=\frac{2}{7}+\frac{14}{25}\)
\(=\frac{50}{175}+\frac{98}{175}=\frac{148}{175}\)
b)
\(\frac{6}{7}+\frac{5}{7}:5-\frac{8}{9}\)
\(=\frac{6}{7}+\frac{5}{30}-\frac{8}{9}\)
\(=\frac{6}{7}+\frac{1}{6}-\frac{8}{9}\)
\(=\frac{36}{42}+\frac{7}{42}-\frac{8}{9}\)
\(=\frac{43}{42}-\frac{8}{9}=\frac{129}{126}-\frac{112}{126}=\frac{17}{126}\)
tk ủng hộ mk nha!!!!!!!!
a) \(\dfrac{2}{7}+\dfrac{7}{7}.\dfrac{14}{25}\)
\(=\dfrac{2}{7}+\dfrac{14}{25}\)
\(=\dfrac{148}{175}\)
b) \(\dfrac{6}{7}+\dfrac{5}{7}:5-\dfrac{8}{9}\)
\(=\dfrac{6}{7}+\dfrac{1}{7}-\dfrac{8}{9}\)
\(=1-\dfrac{8}{9}\)
\(=\dfrac{1}{9}\)
c) \(\left(\dfrac{-5}{28}+1,75+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\dfrac{-69}{20}\)
\(=\dfrac{-12}{23}\)
d) \(70,5-528:\dfrac{15}{2}\)
\(=70,5-\dfrac{352}{5}\)
\(=\dfrac{1}{10}\)
e) \(\dfrac{-7}{25}.\dfrac{39}{-14}.\dfrac{50}{78}\)
\(=\dfrac{\left(-7\right).39.50}{25.\left(-14\right).78}\)
\(=\dfrac{\left(-1\right).2}{\left(-2\right).2}\)
\(=\dfrac{1}{2}\)
a: 14/21=2/3=4/6
60/72=5/6
mà 4<5
nên 14/21<60/72
b: 38/133=2/7=16/56
129/344=3/8=21/56
mà 16<21
nên 38/133<129/344
a) \(\dfrac{-1}{-4}\)=\(\dfrac{1}{4}>0\)
\(\dfrac{3}{-4}< 0\)
\(\Rightarrow\dfrac{1}{4}>\dfrac{3}{-4}hay\dfrac{-1}{-4}>\dfrac{3}{-4}\)
b) Ta có:
\(\dfrac{15}{17}=1-\dfrac{2}{17}\\ \)
\(\dfrac{25}{27}=1-\dfrac{2}{27}\\ \\ \)
Mà \(\dfrac{2}{17}>\dfrac{2}{27}\left(17< 27\right)\)
\(\Rightarrow1-\dfrac{2}{17}< 1-\dfrac{2}{27}\)hay \(\dfrac{15}{17}< \dfrac{25}{27}\)
- Ta có : `6/25=(6.5)/(25.5)=30/125`
`14/125<30/125`
`=> 14/125<6/25`
\(\dfrac{6}{25}=\dfrac{6\cdot5}{25\cdot5}=\dfrac{30}{125}\)
Vậy \(\dfrac{14}{125}< \dfrac{30}{125}\Rightarrow\dfrac{14}{125}< \dfrac{6}{25}\)